AP Calculus AB
AP Calculus AB · Lesson Review · August 26, 2026

Special Trigonometric and Piecewise Limits

Use a small set of special limits and algebraic identities, then compare both sides of absolute-value and piecewise functions.

36-minute edited lesson9 chaptersCaptions + corrected transcript

01 · Watch

Lesson video

VTT captionsSRT captions
0:0001Combine Fractions, Then Rationalize5:4802The Special Limit sin(x)/x8:0303The Reciprocal Limit x/sin(x)10:1204The Cosine Form (1−cos(x))/x18:4205Rewrite Trig Functions Using Sine and Cosine21:0006Scaled-Angle Trigonometric Limits24:3707Absolute Value and One-Sided Limits30:4708Piecewise Limits and Continuity34:3009Choose a Constant for Continuity

Video not loading? Watch directly on YouTube.

02 · Understand

Learning targets

03 · Vocabulary

Key definitions

These ideas connect the algebraic and graphical views of a limit.

01Special trigonometric limit
As x→0x \to 0 in radians, sin⁡xx→1\frac{\sin x}{x} \to 1. Its reciprocal xsin⁡x→1\frac{x}{\sin x} \to 1 as well.
02One-sided limit
The value a function approaches from only the left or only the right of the target input.
03Two-sided limit
A limit that exists only when the left-hand and right-hand limits both exist and are equal.
04Piecewise function
A function defined by different formulas on different parts of its domain.
05Continuity at a point
The left-hand limit, right-hand limit, and function value all agree at that input.
06Conjugate
An expression with the same two terms but the opposite sign between them; it turns a difference of radicals into a difference of squares.

04 · Apply

Representative examples

Look for a form you recognize, then reshape the expression to match it.

Combine + conjugate

Use more than one algebraic technique

  1. lim⁡x→0(1x1+x−1x)\lim_{x \to 0}\left(\frac{1}{x\sqrt{1+x}}-\frac{1}{x}\right)
  2. Use the common denominator x1+xx\sqrt{1+x}.
  3. Rationalize 1−1+x1-\sqrt{1+x} with its conjugate.
  4. Cancel the resulting common factor xx.
  5. −12-\frac{1}{2}
Special limits

Memorize the two reciprocal forms

  1. lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x}=1
  2. lim⁡x→0xsin⁡x=1\lim_{x \to 0}\frac{x}{\sin x}=1
  3. The second result follows by taking the reciprocal of the first limit.
Identity

Reveal sin(x)/x inside a cosine limit

  1. lim⁡x→01−cos⁡xx\lim_{x \to 0}\frac{1-\cos x}{x}
  2. Multiply by 1+cos⁡x1+cos⁡x\frac{1+\cos x}{1+\cos x}.
  3. Use 1−cos⁡2x=sin⁡2x1-\cos^2 x=\sin^2 x.
  4. (sin⁡xx)(sin⁡x1+cos⁡x)⟶1⋅0\begin{aligned}\left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{1+\cos x}\right)\\[-2pt]{}\longrightarrow 1\cdot 0\end{aligned}
  5. =0=0
Rewrite trig

Use sine and cosine as the common language

  1. lim⁡x→0xsec⁡x csc⁡x\lim_{x \to 0}x\sec x\,\csc x
  2. =lim⁡x→0(1cos⁡x)(xsin⁡x)=\lim_{x \to 0}\left(\frac{1}{\cos x}\right)\left(\frac{x}{\sin x}\right)
  3. =1⋅1=1=1\cdot 1=1
Scaled angles

Make the angle and denominator match

  1. lim⁡x→0sin⁡x7x=17⋅1=17\lim_{x \to 0}\frac{\sin x}{7x}=\frac{1}{7}\cdot 1=\frac{1}{7}
  2. lim⁡x→0sin⁡(5x)x=5lim⁡x→0sin⁡(5x)5x=5\begin{aligned}\lim_{x \to 0}\frac{\sin(5x)}{x}&=5\lim_{x \to 0}\frac{\sin(5x)}{5x}\\&=5\end{aligned}
Absolute value

Check the two sides separately

  1. ∣x−3∣=−(x−3)|x-3|=-(x-3) for x<3x<3, so the left-hand limit is 11.
  2. ∣x−3∣=x−3|x-3|=x-3 for x≥3x\ge 3, so the right-hand limit is −1-1.
  3. lim⁡x→3∣x−3∣3−x\lim_{x \to 3}\frac{|x-3|}{3-x}does not exist.
Continuity

Choose the constant that joins two pieces

  1. f(x)={ln⁡x,x≥1,x2+c,x<1.f(x)=\begin{cases}\ln x,&x\ge 1,\\x^2+c,&x<1.\end{cases}
  2. The right-hand limit is ln⁡1=0\ln 1=0.
  3. The left-hand limit is 12+c=1+c1^2+c=1+c.
  4. 1+c=01+c=0, so c=−1c=-1.

05 · Avoid

Common mistakes

Using degrees.

The special trigonometric limits require radian measure.

Mismatching the angle.

For sin⁡(5x)\sin(5x), create 5x5x in the denominator too.

Substituting into the wrong piece.

Choose the formula that applies on the side from which x approaches.

Checking only one side.

A two-sided limit requires agreement from both directions.

Forgetting the outside negative.

cos⁡x−1=−(1−cos⁡x)\cos x-1=-(1-\cos x).

Confusing a limit with f(a)f(a).

Continuity requires both the limit and the function value to match.

06 · Check yourself

Quick check

Try each question before opening its answer.

1What is lim⁡x→0xsin⁡x\lim_{x \to 0}\frac{x}{\sin x}?

1. It is the reciprocal form of the special limit.

2What is lim⁡x→0sin⁡(7x)x\lim_{x \to 0}\frac{\sin(7x)}{x}?

7. Write it as 7⋅sin⁡(7x)7x7\cdot\frac{\sin(7x)}{7x}.

3Why does lim⁡x→01−cos⁡xx\lim_{x \to 0}\frac{1-\cos x}{x} equal 0?

After rationalizing, it becomes (sin⁡xx)(sin⁡x1+cos⁡x)\left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{1+\cos x}\right), which approaches 1⋅01\cdot 0.

4If the left-hand limit is 1 and the right-hand limit is −1, what is the two-sided limit?

It does not exist. The two sides disagree.

5For the piecewise continuity example, what value of cc makes the pieces meet at x=1x=1?

c=−1c=-1. Then the left-hand value 1+c1+c equals ln⁡1=0\ln 1=0.

07 · Revisit

Transcript and captions

Search the corrected transcript for a technique or return to the matching chapter. Caption files are available above for offline viewing.

Download corrected transcript